Slide 1OverviewCAD Model of Microstrip AntennasTank Circuit: complex resonance frequencyResonance Frequency (cont.)Slide 6Slide 7Natural ResponseNatural Response (cont.)Stored EnergyStored Energy (cont.)Slide 12Q of CavityQ of Cavity (Cont.)Slide 15Slide 16Slide 17Input ImpedanceInput Impedance (cont.)Input Impedance (continued)Slide 21Slide 22Reflection CoefficientBandwidthBandwidth (Continued)Slide 26Slide 27Slide 28Slide 29Slide 30Slide 31Complete ModelComplete Model (cont.)Slide 34Slide 35Slide 361Spring 2011Notes 1ECE 6345ECE 6345Prof. David R. JacksonECE Dept.2OverviewOverviewIn this set of notes we discuss the CAD model of the microstrip antenna. Discuss complex resonance frequency Derive formula for Q Derive formula for input impedance Derive formula for impedance bandwidth3CAD Model of Microstrip AntennasCAD Model of Microstrip AntennasLp(probe inductance)tank (RLC) circuitR L C4Tank Circuit: complex resonance frequencyTank Circuit: complex resonance frequencyYurYsuY Y=-su urTransverse Resonance Equation (TRE):1GR�R L C+-VThe complex resonance frequency is denoted as 0.5Resonance Frequency (cont.)Resonance Frequency (cont.)so choose + sign10,GLCw� �( ) ( ) ( )222 2111 042G j Cj Lj LG LCLC jLGjLG L G LCLCwww ww ww� �=- +� �� �=- ++ - + - =� - +=6021 1 1 112 4LjRC R CLCw� �= + -� �� �Resonance Frequency (cont.)Resonance Frequency (cont.)0 0 0jw w w� ��= +Denote:021 114LR CLCw�= -01 12 RCw� ���=� �� �7CLR 01LCw��Forwe haveResonance Frequency (cont.)Resonance Frequency (cont.)021 114LR CLCw�= -01 12 RCw� ���=� �� �01 12 RCw� ���=� �� �8( )0( ) Rej tv t V ew=( )0 0( ) Rej t tv t e ew w� ��-=(Take V = 1)In the time domain:soNatural ResponseNatural ResponseR L C+-V9Natural Response (cont.)Natural Response (cont.)( )v t0tew��-( )00( ) costv t e tww��-�=t0 01 2Tfpw= =� �10Stored EnergyStored Energy( )( )022201( )21cos2EtU t Cv tCe tww��-=�=0 0 01 1VIj L j L j Lw w w= = ��( )( )0 0 0 000 01 1 1( ) Re Re sinj t t j t ti t I e e e e tL j Lw w w www w�� � ��- + -� ��= � =� �� �� �R L C+-v(t)i(t)For the inductor:so(Take V = 1)11Stored Energy (cont.)Stored Energy (cont.)( )( )( )002222002201( )21 1sin21sin2HttU t L i tL e tLCe twwwww��-��-=� ��=� ��� ��=( ) ( )0214tE HU t U t Cew��-< >=< >=Note:R L C+-v(t)i(t)( ) ( ) ( )0212tS E HU t U t U t Cew��-= + =Also, note that12( )0 02 21(0)2t tS SU t C e U ew w�� ��- -= =( )SU tt( ) ( ) ( ) ( )2S E H EU t U t U t U t= + =Stored Energy (cont.)Stored Energy (cont.)Hence13QQ of Cavity of Cavity2STDUQUp� ��� �� �2/STDUQT U Tp� ��� �� �0SAVEDUQPw� ���� �� �AVEDP= energy dissipated per cycle (period T)= average power dissipatedTDUor14QQ of Cavity (Cont.) of Cavity (Cont.)( )0000220 022 20202001 12 2( ) cos1212tttttCe C eQG v t G e tC eG eCGRCwwwwww wwwww����--��-��-��-� � � �� � � �� �= =� � � �< > < >� � � �� � � �� �� ��=� �� �� ��=�=( )( )0002202202coscos12ttte te tewwwww��-��-��-< >� < >� �=� �� �15Q of Cavity (Cont.)Q of Cavity (Cont.)001 R CQ RC RL Lww� ��= = =� ��� �We then haveRecall thatHence0 01 112j LCRCw w� �� ��� +� �� �� �� �01LCw��01 12 RCw� ���=� �� �16Q of Cavity (Cont.)Q of Cavity (Cont.)Hence0 0112jQw w� ��� +� �� �001 1121 112j LCRCLjR Cw ww� �� ��� +� �� �� �� �� �� ��= +� �� �� �� �� �17Q of Cavity (Cont.)Q of Cavity (Cont.)( )020( ) costQv t e tww�� �-� �� ��=0( ) (0)tQS SU t U ew�� �-� �� �=We can thus write18Input ImpedanceInput ImpedanceLjCjGYRLC1LRRCjRLjRRCjRLjCjGZRLC1111RL CThe probe inductance is neglected here.19Input Impedance (cont.)Input Impedance (cont.)( )000 00011RLCRZRj RCLRj Q Qwwww w wwww w=� ��� ��+ -� �� �� �� �� �=� ��� �+ -� �� ��� �� �0 0rfffww= =�11RCLrrRZj Q ff=� �+ -� �� �orThen we haveDefine002fwp��where20Input Impedance (continued)Input Impedance (continued)( )( ) ( )( )21 1111 12 1r rr rr rrrF f ff ff fff� - = -= - +� -( )1 1 2 1RLCrR RZjQF j Q f= �+ + -HenceNote that1rf �for212 21 11 1 1RLC RLC RLCxZ R Xjx x x-= = =+ + +Input Impedance (continued)Input Impedance (continued)11RLCZ RjQF� �=� �+� �RLCRLCZZR=Define( )12 1r rrx QF Q f Q ff� �= = - � -� �� �We then have22Input Impedance (continued)Input Impedance (continued)fRLCXRLCZRLCR0fRLCXRLCR0.51.0- 0.5-11xRLCZ23Reflection CoefficientReflection Coefficient001111 (1 )1 1 (1 )2RLC RLCRLC RLCRLCRLCZ Z ZZ Z ZYjxY jxjxjx- -G= =+ +-- += =+ + +-=+RLCZ00Z R=24BandwidthBandwidth24xxG =+11SWRSBandwidth definition is based on SWR < S0x0x0S0 Sx 1.0f2 f1 S0 Sf 1.0f0 The value S0 is often chosen as 2.0.)25Bandwidth (Continued)Bandwidth (Continued)12012rrfffffBW 1rrx QF Q ff� �� = -� �� �1rrxff Q- =21 0r rxf fQ� �- - =� �� �2242rx xQ Qf� +=soWe can solve for fr in terms of x:Recall that262242rx xQ Qf+ +=20 02242rx xQ Qf+ +=20 02142rx xQ Qf- + +=0, 1rx f� �HenceTo determine correct sign, enforce thatTherefore(So choose the plus sign.)Bandwidth (Continued)Bandwidth (Continued)27110S0011SS-G =+0204xxG =+0xBWQ=Now we need to solve for x0:HencesoandBandwidth (Continued)Bandwidth (Continued)28220 020 014 1x SAx S� �-= =� �+ +� �2 20 04A x A x+ =( )201 4x A A- =0 0200114x SSx-=++ThereforeorBandwidth (Continued)Bandwidth (Continued)2900020002 20 0002111211112( 1) ( 1)124AxASSSSSS SSS=-� �-� �+� �=� �--� �+� �-=+ - -� �-=� �� �� �Bandwidth (Continued)Bandwidth (Continued)The solution is300011SSQBW02S =QBW21Hence0001SxS-=We then haveForBandwidth (Continued)Bandwidth (Continued)3102.0S SWR= =[ ]1110 111320log 9.5 dBSSG = ==-Note:-9.5 f f0 [ ]11dBSBandwidth (Continued)Bandwidth (Continued)32Complete ModelComplete Model0p p pX L Lw w�= �RLCpinZjXZ RL CZ0Lp( )012pXfBWf R� �D� ��� �� �� �� �resffpXR0fX0resf f fD � -(This will be derived in a HW